LeetCode 18 4Sum

LeetCode 18 4Sum

Description

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note: The solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
[-1,  0, 0, 1],
[-2, -1, 1, 2],
[-2,  0, 0, 2]
]

解题思路

求出所有的两个数的和，问题转化为2Sum，二分即可，用Set去重。

代码

class Solution {
public:
struct sum {
sum(int a, int b, int aIdx, int bIdx) : a(a), b(b), aIdx(aIdx), bIdx(bIdx) { c = a + b; }

int a, b, c;
int aIdx, bIdx;

bool operator<(const sum &x) const {
return c < x.c;
}
};

bool checkSum(sum &x, sum &y) const {
return x.aIdx != y.aIdx
&& x.bIdx != y.bIdx
&& x.aIdx != y.bIdx
&& x.bIdx != y.aIdx;
}

int lowerBound(const vector<sum> &sums, int target) {
int l = -1, r = (int) sums.size() - 1, mid = (l + r) >> 1;
while (r - l > 1) {
if (sums[mid].c < target) {
l = mid;
} else {
r = mid;
}
mid = (l + r) >> 1;
}
return r;
}

vector<vector<int>> fourSum(vector<int> &nums, int target) {
vector<sum> sums;
set<vector<int>> ansSet;
vector<vector<int>> ans;
int size = (int) nums.size();
for (int i = 0; i < size; ++i) {
for (int j = i + 1; j < size; ++j) {
sums.emplace_back(sum(nums[i], nums[j], i, j));
}
}
sort(sums.begin(), sums.end());
int sumsSize = (int) sums.size();
for (auto &eachSum:sums) {
int pos = lowerBound(sums, target - eachSum.c);
while (pos < sumsSize && sums[pos].c + eachSum.c == target) {
if (checkSum(sums[pos], eachSum)) {
vector<int> tmp{sums[pos].a, eachSum.a, sums[pos].b, eachSum.b};
sort(tmp.begin(), tmp.end());
ansSet.insert(tmp);
}
++pos;
}
}
for (auto &each:ansSet) {
ans.emplace_back(each);
}
return ans;
}
};